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Answers without the blur. Sign up and see all textbooks for free! Q. 16

Expert-verified Found in: Page 526 ### Geometry

Book edition Student Edition
Author(s) Ray C. Jurgensen, Richard G. Brown, John W. Jurgensen
Pages 227 pages
ISBN 9780395977279 # Given points A, B, and C. Find AB, BC, and AC. Are A, B, and C collinear?If so, which point lies between the other two?16. A(3, 4), B(-3, 0), C(-1, 1)

The values of$AB=2\sqrt{13},\text{}BC=\sqrt{5}\text{,}AC=5$ .

A, B, C are not collinear.

See the step by step solution

## Step-1 – Given

The given points are $A\left(3,4\right)\text{,}B\left(-3,0\right)\text{,}C\left(-1,1\right)$.

## Step-2 – To determine

We have to find the lengths AB, BC and AC. Then have to check if A, B, C are collinear. And determine that which point lies between the other two.

## Step-3 – Calculation

Using the distance formula, the distance between A and B is:

$\begin{array}{l}AB=\sqrt{{\left(-3-3\right)}^{2}+{\left(0-4\right)}^{2}}\\ AB=\sqrt{{\left(-6\right)}^{2}+{\left(-4\right)}^{2}}\\ AB=\sqrt{36+16}\\ AB=\sqrt{52}\\ AB=2\sqrt{13}\end{array}$

Using the distance formula, the distance between B and C is:

$\begin{array}{l}BC=\sqrt{{\left(-1-\left(-3\right)\right)}^{2}+{\left(1-0\right)}^{2}}\\ BC=\sqrt{{\left(-1+3\right)}^{2}+{\left(1-0\right)}^{2}}\\ BC=\sqrt{{\left(2\right)}^{2}+{\left(1\right)}^{2}}\\ BC=\sqrt{4+1}\\ BC=\sqrt{5}\end{array}$

Using the distance formula, the distance between A and C is:

$\begin{array}{l}AC=\sqrt{{\left(-1-3\right)}^{2}+{\left(1-4\right)}^{2}}\\ AC=\sqrt{{\left(-4\right)}^{2}+{\left(-3\right)}^{2}}\\ AC=\sqrt{16+9}\\ AC=\sqrt{25}\\ AC=5\end{array}$

Note that the sum of any two lines is not equal to length of other line.

So, A, B, C are not collinear.

The values of $AB=2\sqrt{13},\text{}BC=\sqrt{5}\text{,}AC=5$. ### Want to see more solutions like these? 